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Phasor circuit

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the voltage Vo using node voltage method specifically.

    2. Relevant equations

    3. The attempt at a solution

    I am able to get the correct answer when finding the equivalent impedance, then determining the current, and then finding the voltage at the first node V1 by subtrating the source from the voltage drop across the inductor. Then I simply use voltage divider to find Vo.

    However, using node voltage at location V1 gets me a completely different answer which means im doing something wrong in my node voltage calculations, but i cannot identify what it is. I am using the TI-84 caclulator to do these calculations.

    Rearranging the eqn to solve for V1 using nodal analysis, V1(1/10j + 1/50 + 1/(30+10j)) = 240/10j= 24j

    V1 = 24j/(1/10j + 1/50 + 1/(30 + 10j))
    So V1 = 203.7735+113.207j which is teh answer provided by the calculator.
    Converting to polar I get 233.108*e^(0.507j) where 0.507 = 0.507*180/pi = 29.04 degrees, which is very differnent according the solution i posted. So where am I going wrong with this calculation? I've tried it three times.
  2. jcsd
  3. Dec 10, 2013 #2


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    Staff: Mentor

    First, V1 is not the same as Vo. Solving for V1 is not solving for Vo.

    Second, your node equation looks suspicious. Can you elaborate how you arrived at it?

    If you want to solve for Vo using just nodal analysis, introduce a second node at the top of the 30 ohm resistor (yes, it's in the middle of a branch and is not an essential node, but it gives you direct access to Vo via node equation). Solve the pair of node equations for Vo.
  4. Dec 10, 2013 #3
    Yes I know, but I cannot possibly get a correct Vo value without a correct V1 value in node voltage.

    So since V1 is incorrect according to the solution i posted, it is the most logical place to look for errors.

    Once a correct V1 is found, then I can find Vo

    And the node voltage for V1 is (V1 - 240)/10j + V1/50 + V1/(30+ 10j) = 0

    And for Vo the node voltage is:
    (Vo - V1)/10j + Vo/30 = 0

    The node voltage uses a KCL (V/Z = I), so my logic for the node voltage is that the voltage across (V1-240) the inductor impedane plus the voltage across the 50 ohm resistor, plus the voltage across the combination of impedanes from the resistor and the second inductor is equivalent to zero by
    Last edited: Dec 10, 2013
  5. Dec 10, 2013 #4


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    Staff: Mentor

    That's not strictly true if you introduce the non-essential node as I suggested. You only need solve the pair of node equations for that node's potential.

    So, check carefully each term of your node equation. Pay attention to signs when terms are moved or reduced by complex operations.
  6. Dec 10, 2013 #5


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    Staff: Mentor

    Okay, those two equations look fine. Use the second one to replace V1 in the first equation and solve for Vo.
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